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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Details: Click on the figure below.
Since AB and CD are diameters of circles O and QSo angle AGD= AKD= 90 => AGKD is cyclicLA.LK=LD.LG=> power of L is the same to both circles O and QSo L is on radical line EF.Similarly for point M
Problem 1321EF is the radical axis of the circles O, Q. Is AB diameter then BH perpendicular in AC and BK perpendicular in AK similar DG, CJ are perpendiculars in AC ,BD respectively .Is <BHC=90=<BJC so BJHC is cyclic or JM.MC=BM.MH. But JM.ME is the power of point M with respect to the circle with center Q and BM.MH is the power of point M with respect to the circle with center O.Therefore the point M belongs to the radical axis EF.Similar The point L belongs to the radical axis EF (<AGD=90=<AKD and AGKD is cyclic GL.LD=AL.LK).APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE